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This is a physically motivated mathematical question. My goal is to compute the integral of the function $$f(x)=\exp\left(-\frac{x^2}{\Omega^2}+i\theta x\right) $$ in the case $\Omega\to +\infty$. This is a simple Gaussian integral, but I'm uncertain about one thing. If I take the limit before computing the integral, then $$\int_{-\infty}^{+\infty} dx \ \lim_{\Omega\to+\infty}f(x)=\int_{-\infty}^{+\infty} dx \ \exp(i\theta x)=\infty. $$ On the other hand, if I take it afterwards I get $$\lim_{\Omega\to +\infty} \int_{-\infty}^{+\infty} dx \ f(x)=\lim_{\Omega\to+\infty}\sqrt{\pi}\Omega \exp\left(-\frac{\Omega^2\theta^2}{4}\right)=0.$$ Now, I know there's a whole theory in mathematics that is about exchanging limits and integrals. But I'm a humble physicist and I care about consistency between the two results. If I know ahead of time that $\Omega$ is arbitrarily large, do I just do away with the squared term as if it never existed, and treat the finite case knowing that the infinite one cannot be derived as its limit?

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    $\begingroup$ No. By taking the limit before the integral, you get $e^{i\theta x}$, which is simply not Lebesgue integrable. So, you just don’t write the symbol $\int_{-\infty}^{\infty}e^{i\theta x}\,dx$. Writing this symbol or saying this is equal to $\infty$ has as much meaning as the equation $\textit{sponge}\ddot{}\text{patrick}\ddot{\smile}\text{bob}=\text{mr krabs}$. So, you just don’t put the limit inside the integral. Once you learn about tempered distributions, you can, with a whole bunch of caveats about notational abuse, regard this as the Fourier transform of $1$, and say it equals $\delta$. $\endgroup$
    – peek-a-boo
    Jul 7, 2023 at 8:32
  • $\begingroup$ @ClaudeLeibovici What do you mean exactly? I know how to take the integral, there is a simple formula for that. $\endgroup$
    – B. Baker
    Jul 7, 2023 at 8:43

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You are right, one has to be careful with limits in the world of distributions, because such expressions can be subjected to various types of regularization.

In the present case, you miscalculated the first limit; indeed, one has by definition $$ \int_\mathbb{R}\mathrm{d}x \lim_{\Omega\to+\infty} f(x) = \int_\mathbb{R}\mathrm{d}x\ e^{i\theta x} = 2\pi\delta(\theta). $$ As for the second limit, it vanishes unless $\theta$ is already zero, so that $$ \lim_{\Omega\to +\infty} \int_\mathbb{R}\mathrm{d}x\ f(x) = \lim_{\Omega\to+\infty} \sqrt{\pi}\,\Omega\,e^{-\frac{1}{4}\Omega^2\theta^2} = \begin{cases} \infty &\mathrm{if}\; \theta = 0 \\ 0 &\mathrm{otherwise} \end{cases} $$ Then, you could show that this expression behaves as a Dirac delta by applying it to a test function $-$ but I'm a bit too lazy today to do it...

Nonetheless, this behaviour could have guessed from the start, since $f(x)$ corresponds to a normal distribution (up to normalization factor) with a shrinking standard deviation and the Dirac delta can be precisely modelled by normal law with a vanishing variance.

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